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A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along the rod. Calculate the potential at the center of curvature of the arc if the potential is assumed to be zero at infinity. (Use any variable or symbol stated above along with the following as necessary: ε0.)

Respuesta :

Lets take a small element of rod on it with very small size

Its total charge on rod is dQ

so we will have potential at the center of curvature due to small point charge is given as

[tex]dV = \frac{kdQ}{R}[/tex]

here given that

radius (R) = a

now the potential due to complete semicircular arc is given as

[tex]V = \int dV[/tex]

[tex]V = \int \frac{k dQ}{a}[/tex]

[tex]V = \frac{k}{a} \int dQ[/tex]

we know that

[tex]k = \frac{1}{4\pi \epsilon_0}[/tex]

now we have

[tex]V = \frac{Q}{4\pi \epsilon_0 a}[/tex]

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